Haloalkanes — Nucleophilic Substitution

Spec Mapping — OCR H432 Module 4.2.2(a)-(b) — Haloalkanes and nucleophilic substitution, covering the polarity of the C-X bond and reactivity of haloalkanes towards nucleophiles; nucleophilic substitution by aqueous hydroxide (→ alcohol), ethanolic cyanide (→ nitrile, with chain extension by one carbon), and ethanolic ammonia (→ primary amine); the SN2 mechanism for primary haloalkanes with curly arrows, transition state, and inversion of configuration; comparison with SN1 (introduced briefly) for tertiary substrates (refer to the official OCR H432 specification document for exact wording).

Haloalkanes are compounds in which one or more hydrogens of an alkane have been replaced by a halogen (F, Cl, Br or I). They are reactive synthetic intermediates because the C-X bond is polar — the carbon carries a partial positive charge that is susceptible to attack by nucleophiles, electron-pair donors that displace the halide leaving group to form a new C-Nu bond. The OCR specification covers three named nucleophilic substitutions: hydroxide (gives alcohol — the most common synthetic interconversion), cyanide (gives nitrile — uniquely extends the chain by one carbon), and ammonia (gives amine — opens the gateway to nitrogen chemistry). All three follow the same SN2 mechanism for primary haloalkanes (concerted backside attack, inversion of configuration, second-order kinetics). Tertiary haloalkanes use a different mechanism (SN1, via a carbocation), which is developed in Lesson 6 alongside elimination. This lesson focuses on the SN2 mechanism with full curly-arrow treatment, the synthetic uses of each named substitution, and the conditions (solvent, temperature, concentration) that determine which product forms.

Key Mechanism: the SN2 mechanism (Substitution Nucleophilic Bimolecular) is a single-step concerted reaction in which the nucleophile attacks the C-X carbon from the opposite side to the leaving group, the C-X bond breaks heterolytically, and the configuration at the carbon inverts (Walden inversion). Rate = k[R-X][Nu] — bimolecular, hence the "2". This mechanism is dominant for primary haloalkanes; tertiary substrates use SN1 (Lesson 6) instead.

The C-X bond — polarity and reactivity

Halogens are more electronegative than carbon, so every C-X bond is polar:

$\overset{\delta+}{\text{C}} - \overset{\delta-}{\text{X}}$Cδ+−Xδ−

Pauling electronegativities: C = 2.55, F = 3.98, Cl = 3.16, Br = 2.96, I = 2.66. The C-X dipole increases F > Cl > Br > I, but the reactivity of the haloalkane towards nucleophilic substitution is the opposite order: R-I > R-Br > R-Cl > R-F. This is because the rate-determining factor is C-X bond strength, not bond polarity:

The C-I bond is weakest and therefore breaks most easily; the C-F bond is strongest and effectively does not undergo nucleophilic substitution at A-Level conditions. This trend reappears in Lesson 5 (hydrolysis of haloalkanes — C-I hydrolyses fastest).

Key Definition: a nucleophile is an electron-pair donor that attacks an electrophilic carbon to form a new bond. Common nucleophiles in A-Level haloalkane chemistry: OH⁻, CN⁻, NH₃ (uses the lone pair on N), H₂O (slow), RO⁻ (alkoxide).

Key Definition: nucleophilic substitution is a reaction in which a nucleophile replaces a leaving group on a saturated carbon. The leaving group takes both bonding electrons with it as X⁻.

Substitution by hydroxide — making alcohols

Heating a haloalkane with aqueous sodium hydroxide (or KOH) under reflux gives the corresponding alcohol:

$R-X + \text{OH}^- \rightarrow R-\text{OH} + X^-$R−X+OH−→R−OH+X−

Worked example for 1-bromobutane → butan-1-ol:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-$CH3​CH2​CH2​CH2​Br+OH−→CH3​CH2​CH2​CH2​OH+Br−

Conditions:

• Aqueous NaOH or KOH (water is the solvent).
• Warm under reflux for ~30 minutes.
• Excess base ensures complete substitution.

The solvent is critical: aqueous NaOH favours substitution (gives alcohol); ethanolic NaOH favours elimination (gives alkene — Lesson 6). The same reagent in different solvents gives different products.

This reaction is also the basis of the hydrolysis test for haloalkanes (Lesson 5): warming the haloalkane with aqueous AgNO₃ produces a halide ion via the same OH⁻ substitution, and the AgX precipitate (white, cream, yellow) identifies the halogen.

flowchart LR
  A[Haloalkane R-X] -->|aq NaOH, reflux| B[Alcohol R-OH + X-]
  A -->|ethanolic NaOH, reflux| C[Alkene + HX - elimination Lesson 6]

Substitution by cyanide — making nitriles (chain extension!)

Heating a haloalkane with ethanolic potassium cyanide replaces the halogen with a cyano group (-CN):

$R-X + \text{CN}^- \rightarrow R-\text{CN} + X^-$R−X+CN−→R−CN+X−

Worked example for bromoethane → propanenitrile:

$\text{CH}_3\text{CH}_2\text{Br} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{Br}^-$CH3​CH2​Br+CN−→CH3​CH2​CN+Br−

Conditions:

• KCN dissolved in ethanol (NOT water — water would let the cyanide partially hydrolyse to ammonia and formate, and would also let OH⁻ compete as a nucleophile).
• Heat under reflux.

Why this reaction is uniquely useful — chain extension

The nitrile product has one more carbon than the starting haloalkane (bromoethane C₂ → propanenitrile C₃). This is one of the very few A-Level reactions that increases the length of the carbon skeleton, and it is therefore invaluable in multi-step synthesis. The "extra" carbon comes from the cyanide nucleophile attaching via its carbon (not via the nitrogen — that would give an isocyanide R-NC, a different and unstable functional group).

Subsequent reactions of the nitrile (Lesson 8 — Organic synthesis techniques):

• Reduction by H₂ / Ni catalyst (or LiAlH₄ in dry ether) → primary amine: $R\text{-CN} + 2\text{H}_2 \rightarrow R\text{-CH}_2\text{NH}_2$R-CN+2H2​→R-CH2​NH2​ (carbon count preserved).
• Acidic hydrolysis by dilute HCl, reflux → carboxylic acid: $R\text{-CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow R\text{-COOH} + \text{NH}_4\text{Cl}$R-CN+2H2​O+HCl→R-COOH+NH4​Cl.

flowchart LR
  A[Haloalkane R-X with n carbons] -->|KCN ethanol reflux| B[Nitrile R-CN with n+1 carbons]
  B -->|H2 / Ni or LiAlH4| C[Primary amine R-CH2-NH2]
  B -->|dilute HCl reflux| D[Carboxylic acid R-COOH]

Synthesis principle: if a question asks you to add one carbon to a chain, always reach for KCN in ethanol. It is one of only three A-Level methods (the others are Friedel-Crafts acylation of benzene and the Grignard reaction, the latter not on A-Level syllabus).

Substitution by ammonia — making amines

Heating a haloalkane with excess ammonia in ethanol in a sealed pressure tube gives a primary amine:

$R-X + 2\text{NH}_3 \rightarrow R-\text{NH}_2 + \text{NH}_4 X$R−X+2NH3​→R−NH2​+NH4​X

Worked example for bromoethane → ethylamine:

$\text{CH}_3\text{CH}_2\text{Br} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Br}$CH3​CH2​Br+2NH3​→CH3​CH2​NH2​+NH4​Br

Conditions:

• Concentrated NH₃ dissolved in ethanol (NH₃ as nucleophile via its nitrogen lone pair).
• Large excess of ammonia (10× or more) to minimise over-alkylation.
• Sealed pressure tube (a "Schlenk tube" or "sealed ampoule") to keep the volatile ammonia in solution — NH₃ boils at -33 °C and would escape from an open flask.
• Warm to ~50-100 °C.

The stoichiometry is 2:1 — one NH₃ acts as the nucleophile (attacking the C-X carbon), the second NH₃ acts as a base (taking the H⁺ that came off the protonated amine intermediate to give NH₄⁺X⁻).

Over-alkylation problem

The primary amine product is itself a nucleophile (lone pair on N) and can react with another haloalkane molecule to give a secondary amine, and so on:

flowchart LR
  A[R-X + NH3] --> B[R-NH2 primary]
  B -->|+ R-X| C[R2NH secondary]
  C -->|+ R-X| D[R3N tertiary]
  D -->|+ R-X| E[R4N+ X- quaternary ammonium salt]

To maximise the primary amine, you use a huge excess of NH₃ so that any haloalkane molecule is statistically more likely to meet an NH₃ molecule than the much-less-abundant primary amine. In practice you still get a mixture and have to separate by distillation; this is one reason why nitrile reduction (Lesson 8 synthesis route) is often preferred for making primary amines.

The SN2 mechanism for primary haloalkanes

OCR requires you to draw the SN2 mechanism for primary haloalkanes with curly arrows, transition state and products. The mechanism has three labelled stages.

Key features of SN2

• One concerted step — no intermediate. Nucleophile-to-C bond formation and C-X bond breaking happen simultaneously in a single transition state.
• Backside attack — nucleophile approaches the C from the side opposite to the leaving group (180° angle).
• Pentacoordinate transition state — the central carbon is partially bonded to five groups (Nu, X and three substituents) and is sp² hybridised at the moment of transition.
• Bimolecular kinetics — rate = k[R-X][Nu], first order in each reactant, second order overall.
• Inversion of configuration (Walden inversion) — the three substituents that stay flip from one side of the carbon to the other, like an umbrella in a strong wind. For a chiral primary carbon (uncommon at A-Level) the stereochemistry inverts.

Reactants HO⁻ → H H C—Br | CH₃ curly arrow: O lone pair → C; C-Br → Br

Transition state ‡ [HO··· C ···Br]‡ H H | CH₃ 5-coordinate C; partial bonds; sp² at C

Products HO—C + Br⁻ H H | CH₃ methanol + bromide; inverted

flowchart LR
  A[HO- attacks C from opposite side to Br] --> B[Pentacoordinate transition state in square brackets with double-dagger]
  B --> C[HO-CH3 + Br- - configuration inverted]